Question

我现在正在编写一个网络抓取工具，我的Python很生疏，所以我只是想知道是否有更短的语法来完成以下操作......

def parse(self, response):
    prc_path = '//span[@class="result-meta"]/span[@class="result-price"]/text()'
    sqf_path = '//span[@class="result-meta"]/span[@class="housing"]/text()'
    loc_path = '//span[@class="result-meta"]/span[@class="result-hood"]/text()'
    prc_resp = response.xpath(prc_path).extract_first()
    sqf_resp = response.xpath(sqf_path).extract_first()
    loc_resp = response.xpath(loc_path).extract_first()
    if sqf_resp and loc_resp:
        yield {
            'prc': response.xpath(prc_path).extract_first(),
            'sqf': response.xpath(sqf_path).extract_first(),
            'loc': response.xpath(loc_path).extract_first()
        }
    elif sqf_resp:
        yield {
            'prc': response.xpath(prc_path).extract_first(),
            'sqf': response.xpath(sqf_path).extract_first()
        }
    else:
        yield {
            'prc': response.xpath(prc_path).extract_first(),
            'loc': response.xpath(loc_path).extract_first()
        }

正如你所看到的，有很多重复，我希望尽可能保持干燥。

Answer 1

您可以创建字典，然后为其添加适当的条目。

result = { 'prc': response.xpath(prc_path).extract_first() }
if sqf_path:
    result['sqf'] = response.xpath(sqf_path).extract_first()
if loc_path:
    result['loc'] = response.xpath(loc_path).extract_first()
yield result

你也可以用dict理解来分解extract_path位。

result = { 'prc': prc_path, 'sqf': sqf_path, 'loc': loc_path }
yield { key : response.xpath(value).extract_first()
          for (key, value) in result.items() if value }

在早期版本的Python中，这将是：

result = { 'prc': prc_path, 'sqf': sqf_path, 'loc': loc_path }
yield dict((key, response.xpath(value).extract_first())
          for (key, value) in result.items() if value)

Answer 2

我选择查找地图：

select date_part(epoch, date) as date
from t2 
where t2.date >= '2014-01-01'
order by date desc

条件语句语法

2 个答案: